Chemistry 2
Lecture 1
Quantum Mechanics in Chemistry
Your lecturers
8am
12pm
Dr Timothy Schmidt
Assoc. Prof. Adam J Bridgeman
Room 457
timothy.schmidt@sydney.edu.au
93512781
Room 222
adam.bridgeman@sydney.edu.au
93512731
Learning outcomes
•Be able to recognize a valid wavefunction in terms of its being
single valued, continuous, and differentiable (where potential is).
•Be able to recognize the Schrödinger equation. Recognize that
atomic orbitals are solutions to this equation which are exact for
Hydrogen.
•Apply the knowledge that solutions to the Schrödinger equation are
the “observable energy levels” of a molecule.
•Use the principle that the mixing between orbitals depends on the
energy difference, and the resonance integral.
•Rationalize differences in orbital energy levels of diatomic
molecules in terms of s-p mixing.
How quantum particles behave
• Uncertainty principle: DxDp > ℏ
• Wave-particle duality
• Need a representation for a quantum
particle: the wavefunction (ψ)
How quantum particles behave
The Wavefunction
• The quantum particle is described by a
function – the wavefunction – which may be
interpreted as a indicative of probability
density
• P(x) = Ψ(x)2
• All the information about the particle is
contained in the wavefunction
The Wavefunction
Must be single valued
Must be continuous
Should be differentiable (where potential is)
The Schrödinger equation
• Observable energy levels of a quantum
particle obey this equation
wavefunction
Hamiltonian operator
Energy
Hydrogen atom
• You already know the solutions to the
Schrödinger equation!
• For hydrogen, they are orbitals…
3s
2s
1s
No nodes
2p
1 node
3p
2 nodes
Hydrogen energy levels
E
2 nodes
3p
(also 3d orbitals…)
3s
2s
1 node
2p
Rydberg constant
energy level
1s
No nodes
 Z
En 
2
n Nuclear charge
Principal quantum number
2
ionization potential
4s
3s
4p
3p
2s
2p
UV (can’t see)
1s
Solutions to Schrödinger equation
• Hamiltonian operates on wavefunction and
gets function back, multiplied by energy.
Ĥ  
•
•
•
•
More nodes, more energy (1s, 2s &c..)
Solutions are observable energy levels.
Lowest energy solution is ground state.
Others are excited states.
Solutions to Schrödinger equation
Ĥ  
Exactly soluble for various model problems
• You know the hydrogen atom (orbitals)
In these lectures we will deal with
• Particle-in-a-box
• Harmonic Oscillator
• Particle-on-a-ring
But first, let’s look at approximate solutions.
Approximate solutions to Schrödinger
equation
• For molecules, we solve the problem
approximately.
• We use known solutions to the Schrödinger
equation to guide our construction of
approximate solutions.
• The simplest problem to solve is H2+
Revision – H2+
• Near each nucleus, electron should behave as a
1s electron.
• At dissociation, 1s orbital will be exact solution
at each nucleus

r
Revision – H2+
• At equilibrium, we have to make the lowest energy possible
using the 1s functions available

r

r
?
Revision – H2+

 = 1sA – 1sB
anti-bonding
1sA
E
1sB
1sB
1sA
 = 1sA + 1sB

bonding
1sA
1sB
Revision – H2

 = 1sA – 1sB
anti-bonding
1sA
E
1sB
1sB
1sA
 = 1sA + 1sB

bonding
1sA
1sB
Revision – He2

 = 1sA – 1sB
anti-bonding
1sA
E
1sB
1sB
1sA
 = 1sA + 1sB

bonding
NOT BOUND!!
1sA
1sB
2nd row homonuclear diatomics
• Now what do we do? So many orbitals!
2p
2p
2s
2s
1s
1s
Interacting orbitals
Orbitals can interact and combine to make new approximate
solutions to the Schrödinger equation. There are two
considerations:
1. Orbitals interact inversely proportionally to their energy
difference. Orbitals of the same energy interact completely,
yielding completely mixed linear combinations.
2. The extent of orbital mixing is given by the integral
total energy operator (Hamiltonian)
ˆ  d

H
 2 1
integral over all space
orbital wavefunctions
Interacting orbitals
1. Orbitals interact inversely proportionally to their energy
difference. Orbitals of the same energy interact completely,
yielding completely mixed linear combinations.

1
( 2 sA  2 sB )
2
2p
2p
2s
2s
 2 sA
1s

1
( 2 sA  2 sB )
2
 2 sB
1s
Interacting orbitals
1. The extent of orbital mixing is given by the integral
ˆ  d  something

H
 2 1
2p
2s
2p
 2 s
 2 p
2s
The 2s orbital on one atom can interact with the 2p from the
other atom, but since they have different energies this is a
smaller interaction than the 2s-2s interaction. We will deal
with this later.
1s
1s
Interacting orbitals
1. The extent of orbital mixing is given by the integral
ˆ  d  0

H
 2 1
cancels
2p
2s
2p
 2 s
 2 p
2s
There is no net interaction between these orbitals.
The positive-positive term is cancelled by the positive-negative term
1s
1s
(First year) MO diagram
Orbitals interact most with the corresponding orbital on the other atom
to make perfectly mixed linear combinations. (we ignore core).
2p
2p
2s
2s
Molecular Orbital Theory - Revision
•Molecular orbitals may be classified according to their symmetry
•Looking end-on at a diatomic molecule, a molecular orbital may
resemble an s-orbital, or a p-orbital.
•Those without a node in the plane containing both nuclei resemble an
s-orbital and are denoted -orbitals.
•Those with a node in the plane containing both nuclei resemble a porbital and are denoted -orbitals.


Molecular Orbital Theory - Revision
•Molecular orbitals may be classified according to their contribution to
bonding
•Those without a node between the nuclei resemble are bonding.
•Those with a node between the nuclei resemble are anti-bonding,
denoted with an asterisk, e.g. *.
*
*
Molecular Orbital Theory - Revision
• Can predict bond strengths qualitatively
2 p *
2 p *
2 p
2 p
2 s *
N2 Bond Order = 3
diamagnetic
2s
More refined MO diagram
 orbitals can now interact
2 p *
2 p *

2 p
2 p
2 s *
2s

More refined MO diagram
* orbitals can interact
2 p *
*
2 p *

2 p
2 p
2 s *
*
2s

More refined MO diagram
 orbitals do not interact
*
2 p *
2 p *
2 p
2 p
2 s *
*


*
2s

More refined MO diagram
sp mixing
*
2 p *
2 p *
*
2 p

2 p
2 s
*


*
2s

This new interaction energy
Depends on the energy
spacing between the 2s
and the 2p
Smallest energy gap,
and thus largest
mixing between 2s
and 2p is for Boron.
sp mixing
Largest energy gap, and
thus smallest mixing
between 2s and 2p is for
Fluorine.
2p
2s
c.f.
 Z 2
En 
n2
sp mixing
*
*
*
*




*
weakly bound




*
*
diamagnetic



B2
*
*

*
paramagnetic

Be2
*


*
C2
N2
sp mixing in N2
sp mixing in N2
-bonding orbital
Linear combinations of lone pairs
:N≡N:
Summary
• A valid wavefunction is single valued, continuous, and
differentiable (where potential is).
• Solving the Schrödinger equation gives the observable energy
levels and the wavefunctions describing a sysyem:
• Atomic orbitals are solutions to the Schrödinger equation for
atoms and are exact for hydrogen.
• Molecular orbitals are solutions to the Schrödinger equation for
molecules and are exact for H2+.
• Molecular orbitals are made by combining atomic orbitals and
are labelled by their symmetry with σ and π used for diatomics
• Mixing between atomic orbitals depends on the energy
difference, and the resonance integral.
• The orbital energy levels of diatomic molecules are affected by
the extent of s-p mixing.
Next lecture
• Particle in a box approximation
– solving the Schrödinger equation.
Week 10 tutorials
• Particle in a box approximation
– you solve the Schrödinger equation.
Practice Questions
1. Which of the wavefunctions (a) – (d) is
acceptable as a solution to the Schrödinger
equation?
2. Why is s-p mixing more important in Li2
than in F2?
3. The ionization energy of NO is 9.25 eV
and corresponds to removal of an
antibonding electron
(a) Why does ionization of an antibonding
electron require energy?
(b) Predict the effect of ionization on the
bond length and vibrational frequency
of NO
(c) The ionization energies of N2, NO, CO
and O2 are 15.6, 9.25, 14.0 and 12.1
eV respectively. Explain.